\(BA=\sqrt{12^2+8^2}=4\sqrt{13}=4BC\)
Do \(C\in\) tia BA \(\Rightarrow\overrightarrow{BC}=\frac{1}{4}\overrightarrow{BA}\)
Gọi \(C\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{BC}=\left(x+11;y+25\right)\\\overrightarrow{BA}=\left(12;8\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+11=\frac{1}{4}.12\\y+25=\frac{1}{4}.8\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-8\\y=-23\end{matrix}\right.\) \(\Rightarrow C\left(-8;-23\right)\)